Question: Graph this system of equations and solve. $x+y = 3$ $-12x-2y = 4$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Convert the first equation, $x+y = 3$ , to slope-intercept form. $y = - x + 3$ The y-intercept for the first equation is $3$ , so the first line must pass through the point $(0, 3)$ The slope for the first equation is $-1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $1$ position to the right. $1$ position to the right. $1$ position down from $(0, 3)$ is $(1, 2)$ Graph the blue line so it passes through $(0, 3)$ and $(1, 2)$ Convert the second equation, $-12x-2y = 4$ , to slope-intercept form. $y = -6 x - 2$ The y-intercept for the second equation is $-2$ , so the second line must pass through the point $(0, -2)$ The slope for the second equation is $-6$ . Remember that the slope tells you rise over run. So in this case for every $6$ positions you move down (because it's negative) $1$ position to the right. $6$ positions down from $(0, -2)$ is $(1, -8)$ Graph the green line so it passes through $(0, -2)$ and $(1, -8)$ The solution is the point where the two lines intersect. The lines intersect at $(-1, 4)$.